Three rods of different materials are connected and placed between rigid supports at A and D, as shown in Figure P5.66/67. Properties for each of the three rods are given below. The bars are initially unstressed when the structure is assembled at 70°F. After the temperature has been increased to 250°F, determine:
(a) the normal stresses in the three rods.
(b) the force exerted on the rigid supports.
(c) the deflections of joints B and C relative to rigid support A.

Question

[latex]\begin{array}{|l|l|l|l|}\hline ext { Aluminum (1) } & ext { Cast Iron (2) } & ext { Bronze (3) } \L_{1}=10 ext { in. } & L_{2}=5 ext { in. } & L_{3}=7 ext { in. } \A_{1}=0.8 ext { in. }^{2} & A_{2}=1.8 ext { in. }^{2} & A_{3}=0.6 ext { in. }^{2} \E_{1}=10,000 ksi & E_{2}=22,500 ksi & E_{3}=15,000 ksi \\alpha_{1}=12.5 imes 10^{-6} /{ }^{\circ} F & \alpha_{2}=7.5 imes 10^{-6} /{ }^{-6} F & \alpha_{3}=9.4 imes 10^{-6} /{ }^{\circ} F \\hline\end{array}[/latex]

Accepted Answer

Equilibrium
Consider a FBD at joint B. Assume that both internal axial forces will be tension.

[latex]\Sigma F_{x}=-F_{1}+F_{2}=0 \quad herefore F_{1}=F_{2}[/latex] (a)

Similarly, consider a FBD at joint C. Assume that both internal axial forces will be tension.

[latex]\Sigma F_{x}=-F_{2}+F_{3}=0 \quad herefore F_{3}=F_{2}[/latex] (b)

Geometry of Deformations Relationship

[latex]\delta_{1}+\delta_{2}+\delta_{3}=0[/latex] (c)

Force-Temperature-Deformation Relationships

[latex]\delta_{1}=\frac{F_{1} L_{1}}{A_{1} E_{1}}+\alpha_{1} \Delta T_{1} L_{1} \quad \delta_{2}=\frac{F_{2} L_{2}}{A_{2} E_{2}}+\alpha_{2} \Delta T_{2} L_{2} \quad \delta_{3}=\frac{F_{3} L_{3}}{A_{3} E_{3}}+\alpha_{3} \Delta T_{3} L_{3}[/latex] (d)

Compatibility Equation

[latex]\frac{F_{1} L_{1}}{A_{1} E_{1}}+\alpha_{1} \Delta T_{1} L_{1}+\frac{F_{2} L_{2}}{A_{2} E_{2}}+\alpha_{2} \Delta T_{2} L_{2}+\frac{F_{3} L_{3}}{A_{3} E_{3}}+\alpha_{3} \Delta T_{3} L_{3}=0[/latex] (e)

Solve the Equations

From Eq. (a), [latex]F_{1}=F_{2}[/latex], and from Eq. (b), [latex]F_{3}=F_{2}[/latex]. The temperature change is the same for all members;
therefore, [latex]\Delta T_{1}=\Delta T_{2}=\Delta T_{3}=\Delta T[/latex]. Eq. (e) then can be written as:

[latex]\frac{\left(F_{2} ight) L_{1}}{A_{1} E_{1}}+\alpha_{1} \Delta T L_{1}+\frac{F_{2} L_{2}}{A_{2} E_{2}}+\alpha_{2} \Delta T L_{2}+\frac{\left(F_{2} ight) L_{3}}{A_{3} E_{3}}+\alpha_{3} \Delta T L_{3}=0[/latex]

Solving for [latex]F_{2}[/latex] :

[latex]\begin{gathered}\frac{F_{2} L_{1}}{A_{1} E_{1}}+\frac{F_{2} L_{2}}{A_{2} E_{2}}+\frac{F_{3} L_{3}}{A_{3} E_{3}}=-\alpha_{1} \Delta T L_{1}-\alpha_{2} \Delta T L_{2}-\alpha_{3} \Delta T L_{3} \F_{2}\left[\frac{L_{1}}{A_{1} E_{1}}+\frac{L_{2}}{A_{2} E_{2}}+\frac{L_{3}}{A_{3} E_{3}} ight]=-\Delta T\left[\alpha_{1} L_{1}+\alpha_{2} L_{2}+\alpha_{3} L_{3} ight] \F_{2}=-\frac{\Delta T\left[\alpha_{1} L_{1}+\alpha_{2} L_{2}+\alpha_{3} L_{3} ight]}{\frac{L_{1}}{A_{1} E_{1}}+\frac{L_{2}}{A_{2} E_{2}}+\frac{L_{3}}{A_{3} E_{3}}} (f)\end{gathered}[/latex]

Substitute the problem data along with ΔT = +180°F into Eq. (f) and calculate [latex]F_{1}[/latex] = −19.1025 kips.
From Eq. (a), [latex]F_{1}[/latex] = −19.1025 kips and from Eq. (b), [latex]F_{3}[/latex] = −19.1025 kips.

(a) Normal Stresses
The normal stresses in each rod can now be calculated:

[latex]\begin{aligned}&\sigma_{1}=\frac{F_{1}}{A_{1}}=\frac{-19.1025 kips }{0.8 in. ^{2}}=-23.878 ksi =23.9 ksi ( C ) \&\sigma_{2}=\frac{F_{2}}{A_{2}}=\frac{-19.1025 kips }{1.8 in .^{2}}=-10.613 ksi =10.61 ksi ( C ) \&\sigma_{3}=\frac{F_{3}}{A_{3}}=\frac{-19.1025 kips }{0.6 in .^{2}}=-31.838 ksi =31.8 ksi ( C )\end{aligned}[/latex]

(b) Force on Rigid Supports
The force exerted on the rigid supports is equal to the internal axial force:

[latex]R_{A}=R_{D}=19.10 kips[/latex]

(c) Deflection of Joints B and C
The deflection of joint B is equal to the deformation (i.e., contraction in this instance) of rod (1). The deformation of rod (1) is given by:

[latex]\begin{aligned}\delta_{1} &=\frac{F_{1} L_{1}}{A_{1} E_{1}}+\alpha_{1} \Delta T_{1} L_{1} \&=\frac{(-19.1025 kips )(10 in .)}{\left(0.8 in .{ }^{2} ight)(10,000 ksi )}+\left(12.5 imes 10^{-6} /{ }^{\circ} F ight)\left(180^{\circ} F ight)(10 in .)=-0.001378 in .\end{aligned}[/latex]

The deflection of joint B is thus:

[latex]u_{B}=\delta_{1}=0.001378 ext { in. } \leftarrow[/latex]

The deformation of rod (2) is given by:

[latex]\begin{aligned}\delta_{2} &=\frac{F_{2} L_{2}}{A_{2} E_{2}}+\alpha_{2} \Delta T_{2} L_{2} \&=\frac{(-19.1025 kips )(5 in .)}{\left(1.8 in .^{2} ight)(22,500 ksi )}+\left(7.5 imes 10^{-6} /{ }^{\circ} F ight)\left(180^{\circ} F ight)(5 in .)=0.004392 in .\end{aligned}[/latex]

The deflection of joint C is:

[latex]u_{C}=u_{B}+\delta_{2}=-0.001378 in .+0.004392 in .=0.00301 in . ightarrow[/latex]

Question 5.66: Three rods of different materials are connected and placed b...

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